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Chapter 3: Problem 66

Partitioning large square matrices can sometimes make their inverses easier to compute, particularly if the blocks have a nice form. In Exercises \(64-68\), verify by block multiplication that the inverse of a matrix, if partitioned as shown, is as claimed. (Assume that all inverses exist as needed.) $$\left[\begin{array}{cc}I & B \\\C & I\end{array}\right]^{-1}=\left[\begin{array}{cc}(I-B C)^{-1} & -(I-B C)^{-1} B \\\ -C(I-B C)^{-1} & I+C(I-B C)^{-1} B\end{array}\right]$$

Short Answer

Expert verified
Verified: The inverse is correct by block multiplication.

Step by step solution

01

Define the Problem

The goal is to verify, by block multiplication, the claimed inverse of a partitioned matrix. We have: \[ A = \begin{bmatrix} I & B \ C & I \end{bmatrix} \] The claimed inverse is: \[ A^{-1} = \begin{bmatrix} (I-BC)^{-1} & -(I-BC)^{-1}B \ -C(I-BC)^{-1} & I + C(I-BC)^{-1}B \end{bmatrix} \] We need to show that \( A \times A^{-1} = I \), the identity matrix.
02

Calculate the Block Multiplication

Perform block multiplication between matrices \( A \) and \( A^{-1} \):\[A \times A^{-1} = \begin{bmatrix} I & B \ C & I \end{bmatrix} \times \begin{bmatrix} (I-BC)^{-1} & -(I-BC)^{-1}B \ -C(I-BC)^{-1} & I+C(I-BC)^{-1}B \end{bmatrix}\]Carry out the multiplication for each block position:
03

Calculate Top-Left Block

Multiply to find the top-left block:\[I \cdot (I-BC)^{-1} + B \cdot (-C(I-BC)^{-1}) = (I-BC)^{-1} - B \cdot C \cdot (I-BC)^{-1}\]Using the distributive property:\[= (I-BC)(I-BC)^{-1} = I\]
04

Calculate Top-Right Block

Multiply to find the top-right block:\[I \cdot (-(I-BC)^{-1}B) + B \cdot (I+C(I-BC)^{-1}B) = -(I-BC)^{-1}B + B + B(C(I-BC)^{-1}B)\]Simplify:\[= -(I-BC)^{-1}B + (BC)(I-BC)^{-1}B = 0\]
05

Calculate Bottom-Left Block

Multiply to find the bottom-left block:\[C \cdot (I-BC)^{-1} + I \cdot (-C(I-BC)^{-1}) = C(I-BC)^{-1} - C(I-BC)^{-1} = 0\]
06

Calculate Bottom-Right Block

Multiply to find the bottom-right block:\[C \cdot (-(I-BC)^{-1}B) + I \cdot (I + C(I-BC)^{-1}B)\]Simplifying:\[= -C(I-BC)^{-1}B + I + C(I-BC)^{-1}B = I\]
07

Validate the Result

After performing multiplications for each block of \( A \times A^{-1} \), we have:\[A \times A^{-1} = \begin{bmatrix} I & 0 \ 0 & I \end{bmatrix} = I\]Thus, the claimed inverse is verified correct by block multiplication.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

block multiplication
Block multiplication is a technique used in matrix algebra where we multiply subdivided or "blocked" matrices systematically, much like traditional matrix multiplication. Instead of multiplying individual numbers, we work with submatrices or "blocks." Here is how it simplifies the computation:
  • Each "block" is treated as a single entity, which can help organize and simplify matrix operations.
  • By focusing on block-level operations, computations that are complex if done element-by-element can become more manageable.
  • This approach is especially useful for larger matrices where direct element-wise multiplication would be cumbersome.
In our exercise, we multiply blocks such as \(I\), \(B\), \(C\), and combinations like \((I-BC)^{-1}\). This allows us to structure the process efficiently by looking at the product of each corresponding block.
partitioned matrices
Partitioned matrices refer to a technique in matrix algebra where a larger matrix is divided into smaller submatrices or "blocks". It's often used to simplify the analysis or computation of matrix-related problems. Here are some key points:
  • Partitioning is useful for handling large matrices by breaking them into smaller, more manageable sections.
  • Each partition can be subjected to operations just like a regular matrix, which simplifies complex matrix calculations.
  • This technique allows specific blocks to be targeted in calculations, such as finding inverses or solving linear equations.
In this exercise, the matrix \(A\) is partitioned into four blocks: \(I\), \(B\), \(C\), and another \(I\). Each of these blocks is individually operated upon to verify the inverse matrix, demonstrating the power of partitioned matrices in solving complex matrix operations.
linear algebra
Linear algebra is a branch of mathematics focused on vectors, vector spaces, and linear transformations. It is essential in understanding concepts such as matrices and their operations. Here's why linear algebra is vital:
  • It provides the tools to solve systems of linear equations, which are foundational in various scientific fields.
  • Linear algebra serves as the groundwork for more complex subjects like machine learning and quantum physics.
  • Understanding matrices and their properties, such as rank, determinant, and inverse, is critical in applications like computer graphics and data analysis.
The exercise in question leverages linear algebra to validate the inverse of a matrix using block multiplication, highlighting just one of its many applications. By employing concepts from linear algebra, we can solve problems involving large and complex matrices efficiently.
inverse of a matrix
The inverse of a matrix is the matrix that, when multiplied by the original matrix, yields the identity matrix. It plays a crucial role in solving linear systems and other mathematical operations. Here are some critical points about matrix inverses:
  • A matrix must be square (same number of rows and columns) and invertible to have an inverse.
  • The identity matrix, which acts like the number 1 in matrix operations, is central to verifying an inverse. When multiplied by a matrix, the inverse results in the identity matrix \(I\).
  • Inverses are used in various applications like computing solutions to linear systems or transforming coordinate spaces.
In our specific exercise, the inverse of a partitioned matrix is determined and verified through block multiplication. This demonstrates how the calculation of an inverse, even for complex matrices, can be managed through block methods, providing a clearer path to understanding and utilizing the concept of matrix inverses in linear algebra.

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Most popular questions from this chapter

Draw a digraph that has the given adjacency matrix. $$\left[\begin{array}{ccccc} 0 & 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 \end{array}\right]$$

Prove that \(T: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}\) is a linear transformation if and only if \\[ T\left(c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}\right)=c_{1} T\left(\mathbf{v}_{1}\right)+c_{2} T\left(\mathbf{v}_{2}\right) \\] for all \(\mathbf{v}_{1}, \mathbf{v}_{2}\), in \(\mathbb{R}^{n}\) and scalars \(c_{1}, c_{2}\)

Let \(\ell\) be a line through the origin in \(\mathbb{R}^{2}, P_{\ell}\) the linear transformation that projects a vector onto \(\ell\), and \(F_{\ell}\) the transformation that reflects a vector in \(\ell\) (a) Draw diagrams to show that \(F_{\ell}\) is linear. (b) Figure 3.14 suggests a way to find the matrix of \(F_{c}\) using the fact that the diagonals of a parallelogram bisect each other. Prove that \(F_{\ell}(\mathbf{x})=2 P_{\ell}(\mathbf{x})-\mathbf{x}\) and use this result to show that the standard matrix of \(F_{\ell}\) is \\[ \frac{1}{d_{1}^{2}+d_{2}^{2}}\left[\begin{array}{cc} d_{1}^{2}-d_{2}^{2} & 2 d_{1} d_{2} \\ 2 d_{1} d_{2} & -d_{1}^{2}+d_{2}^{2} \end{array}\right] \\] (where the direction vector of \(\ell\) is \(\mathrm{d}=\left[\begin{array}{l}d_{1} \\ d_{2}\end{array}\right]\) ). (c) If the angle between \(\ell\) and the positive \(x\) -axis is \(\theta\) show that the matrix of \(F_{\ell}\) is \\[ \left[\begin{array}{rr} \cos 2 \theta & \sin 2 \theta \\ \sin 2 \theta & -\cos 2 \theta \end{array}\right] \\]

Find the standard matrix of the composite transformation from \(\mathbb{R}^{2}\) to \(\mathbb{R}^{2}\). Clockwise rotation through \(45^{\circ},\) followed by projection onto the \(y\) -axis, followed by clockwise rotation through \(45^{\circ}\).

Let \(A\) be an \(n \times n\) matrix, \(A \geq 0 .\) Suppose that \(A \mathbf{x} < \mathbf{x}\) for some \(\mathbf{x}\) in \(\mathbb{R}^{n}, \mathbf{x} \geq 0 .\) Prove that \(\mathbf{x} > \mathbf{0}\).
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